suppose that 4.00 mol of N 2 O 4(g) is injectedinto a 1.00L container at 55 o C
ID: 683216 • Letter: S
Question
suppose that 4.00 mol of N2O4(g) is injectedinto a 1.00L container at 55oC and the followingreaction proceeds toward equilibrium:2NO2(g) <-----------> N2O4(g) Ke = 1.15 at55oC
what is the equilibrium concentration of nitrogen dioxide?
this is what i did
I set up an ICE Chart
2NO2(g) <-----------> N2O4(g) Ke = 1.15 at55oC
I (mol/L) 4.00 0
C (mol/L) -2x +x
E (mol/L) 4.00-2x x
Ke = [N2O4(g)]/[NO2(g) ]2
1.15 =x/(4.00-2x)2 (4.00-2x)2 = 4x2 -16x + 16
(1.15)(4x2 -16x + 16) = x
4.6x2 -18.4x + 18.4 = x ( x goes to the otherside)
4.6x2 -19.4x + 18.4 = 0
i used the quadratic formula
x = -(-19.4)±(-19.4)2 - 4(4.6)(18.4) /2(4.6)
x = (19.4 ± 6.15) / 9.2
x = 2.77(invalid because it would give a negative conc'n)or x = 1.44 (i used this value)
next i used the value of x to find the conc'n of nitrogendioxide
[NO2(g) ] = 4.00 - 2x
= 4.00 - 2(1.44)
= 1.12 mol/L
the back of the book says its [NO2(g) ] = 1.66mol/L
Did i do something wrong? This exact same question was just on theprevious page the only thing different about that was that theinitial concentration of N2O4(g) was0.650mol/L. I was able to get the correct answer for that question,but by following the same method i could not get the answer to thisproblem?I did recheck my numbers, but i can't find the mistake...
Explanation / Answer
The 4.00 mol/L actually belongs on the right side of the ICE table.Since the question states that "suppose that 4.00 mol ofN2O4(g)..." 2NO2(g) N2O4(g) Ke = 1.15 at55oC I (mol/L) 0 4.00 C (mol/L) +2x -x E (mol/L) 2x 4.00-x Try that. Otherwise, the steps you did were correct.Related Questions
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